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<p>When the velocity is zero, the body reaches the maximum height. In (<a href="" class="xref" data-knowl="./knowl/eq2_18_1.html" title="Equation 2.5.2">(2.5.2)</a>), setting <span class="process-math">\(v=0\text{,}\)</span> one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_18_1.html ./knowl/eq2_20.html">
\begin{equation*}
0=\left(v_0+\frac{mg}{k}\right) e^{-\frac{k}{m}t}-\frac{mg}{k},
\end{equation*}
</div>
<p class="continuation">which leads to</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_18_1.html ./knowl/eq2_20.html">
\begin{equation*}
t=\frac{m}{k} \ln \left(\frac{v_0 k}{mg}+1\right).
\end{equation*}
</div>
<p class="continuation">This is the time when the body reaches the maximum height. And the maximum height is obtained by taking this time into (<a href="" class="xref" data-knowl="./knowl/eq2_20.html" title="Equation 2.5.4">(2.5.4)</a>), one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_18_1.html ./knowl/eq2_20.html">
\begin{equation*}
x=\frac{m v_0}{k}-\frac{m^2 g}{k^2} \ln \left(\frac{v_0 k}{mg}+1 \right).
\end{equation*}
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